The reactions taking place in the dry cell are: Anode: Zn -> Zn^2+ + 2e^– ; Cathode: 2MnO2 + 2NH4^+ — Electrochemistry Chemistry Question
Question
The reactions taking place in the dry cell are: Anode: Zn -> Zn^2+ + 2e^– ; Cathode: 2MnO2 + 2NH4^+ + 2e^– -> Mn2O3 + 2NH3 + H2O. The minimum mass of reactants, if a dry cell is to generate 0.25A for 9.65 h, are (Mn = 55, Zn = 65.4) (neglect any other chemical reactions occurring in the cell)
💡 Solution & Explanation
Total charge Q = I × t = 0.25 A × 9.65 × 3600 s = 8685 C. The number of Faradays passed = 8685 / 96500 = 0.09 F. From the stoichiometry of the half-reactions, 2 moles of electrons consume exactly 1 mole of Zn, 2 moles of MnO2, and 2 moles of NH4^+. Thus, for 0.09 F: Moles of Zn consumed = 0.09 / 2 = 0.045 mol, so mass of Zn = 0.045 × 65.4 = 2.943 g. Moles of MnO2 consumed = 0.09 mol, so mass of MnO2 = 0.09 × (55 + 32) = 0.09 × 87 = 7.83 g. Moles of NH4^+ consumed = 0.09 mol, so mass of NH4^+ = 0.09 × (14 + 4) = 0.09 × 18 = 1.62 g. Therefore, correct answer is A,B,C.